g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

Question

4 years ago

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g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

erpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion.

Physics

2 answers:

grin007 [14] · Answer 1 · 2021-11-29T23:25:47+03:00

<h2>Answer:</h2><h2></h2>

(a) 422.2 x 10⁻⁷m

(b) 4.6 x 10⁵ Hz

<h2>Explanation:</h2>

The magnetic force acting perpendicular to the velocity of the electron will cause a circular motion so that the centripetal force, $F_{c}$ , of the electron is equal to the Lorentz's force, $F_{l}$ . i.e

$F_{c}$ = $F_{l}$ ---------------(i)

Where;

$F_{c}$ = $\frac{mv^2}{r}$

$F_{l}$ = qvB

Equation (i) then becomes;

$\frac{mv^2}{r}$ = qvB ------------------(ii)

Where;

m = mass of the electron

v = linear velocity of the electron

r = radius of the electron's path

q = charge of the electron

B = magnetic field.

Make r subject of the formula in equation (ii)

r = $\frac{mv^2}{qvB}$

r = $\frac{mv}{qB}$ -----------(iii)

From the question;

v = 121m/s

B = 1.63 x 10⁻⁵ T

q = 1.6 x 10⁻¹⁹ C (known constant)

m = 9.1 x 10⁻³¹kg

Substitute these values into equation (iii) as follows;

r = $\frac{9.1*10^{-31} * 121}{1.6*10^{-19}*1.63*10^{-5}}$

r = 422.2 x 10⁻⁷m

Therefore, the radius of the electron's path is 422.2 x 10⁻⁷m

(ii) The frequency, f, of the motion which is also called the cyclotron frequency - the number of cycles the electron completes around the path every second - is given by;

f = $\frac{v}{2\pi r}$

Substitute the values of v and r into the equation as follows;

f = $\frac{121}{2\pi (422.2*10^{-7})}$

Take $\pi$ = 3.142

f = $\frac{121}{2(3.142) (422.2*10^{-7})}$

f = 4.6 x 10⁵ Hz

Therefore, the frequency of the motion is 4.6 x 10⁵ Hz

kolbaska11 [484] · Answer 2 · 2021-11-29T21:30:16+03:00

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.